# College Math Teaching

## January 12, 2012

### So you want to take a course in complex variables

Ok, what should you have at your fingertips prior to taking such a course?

I consider the following to be minimal prerequisites:

Basic calculus

1. limits (epsilon-delta, 2-d limits)

2. limit definition of the derivative

3. basic calculus differentiation and integration formulas:
chain rule, product rule, quotient rule, integration and differentiation of polynomials, log, exponentials, basic trig functions, hyperbolic trig functions, inverse trig functions.

4. Fundamental Theorem of calculus.

5. Sequences (convergence)

6. Series: geometric series test, ratio test, comparison tests

7. Power series: interval of convergence, absolute convergence

8. Power series: term by term differentiation, term by term integrals

9. Taylor/Power series for 1/(1-x), sin(x), cos(x), exp(x)

Multi-variable calculus

1. partial derivatives

3. parametrized curves

4. polar coordinates

5. line and path integrals

6. conservative vector fields

7. Green’s Theorem (for integration of a closed loop in a plane)

The challenge
Some of complex variables will look “just like calculus”. And, some of the calculations WILL be “just like calculus; for example it will turn out if $\delta$ is any piecewise smooth curve running from $z_1$ to $z_2$ then $\int_{\delta} e^z dz = e^{z_2} - e^{z_1}$. But in many cases, the similarity vanishes and more care must be taken.

You will learn many things such as:
1. The complex function $sin(z)$ is unbounded!

2. No non-constant everywhere differentiable function is bounded; compare that to $f(x) = \frac{1}{1+x^2}$ in calculus.

3. Integrals can have some strange properties. For example, if $\delta$ is the unit circle taken once around in the standard direction, $\int_{\delta} Log(z) dz$ depends on where one chooses to start and stop, even if the start and stop points are the same!

4. You’ll come to understand why the Taylor series (expanded about $x = 0$) for $\frac{1}{1+x^2}$ has radius of convergence equal to one…it isn’t just an artifact of the trick used to calculate the series.

5. You’ll come to understand that being differentiable on an open disk is a very strong condition for complex functions; in particular being differentiable on an open disk means being INFINITELY differentiable on that open set (compare to $f(x) = x^{4/3}$ which has one derivative but NOT two derivatives at $x = 0$

There is much more, of course.